ZEROSS

\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)

Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)

Ta có: \(b^2=a\cdot c\)

=>\(\frac{a}{b}=\frac{b}{c}\) (1)

Ta có: \(c^2=bd\)

=>\(\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

a: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3-\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3-d^3}=\frac{d^3k^3\left(k^6+k^3-1\right)}{d^3\left(k^6+k^3-1\right)}=k^3\)

\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

Do đó: \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)

b: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

\(=\frac{\left(dk^3\right)^3+\left(dk^2\right)^3+\left(dk\right)^3}{\left(dk^2\right)^3+\left(dk\right)^3+d^3}=\frac{d^3k^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Ta có: \(b^2=ac;c^2=bd\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)

Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}=l\) ta có:

\(\left\{{}\begin{matrix}\left(\dfrac{a+b-c}{b+c-d}\right)^3=l^3\\\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=l^3\end{matrix}\right.\Rightarrowđpcm\)

\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)

Thay vào r tính

Giúp mik với chiều đi học rùi khó quá

\(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

= \(\dfrac{a^3+a.c.b+b.d.c}{a.c.b+b.d.c+d^3}\)

= \(\dfrac{a^3}{d^3}=\dfrac{a}{d}\)

Đề có sai k bạn ??

to bi nham

Đề bị lỗi hiển thị hay sao ấy, mình không nhìn thấy BĐT/ đẳng thức bạn muốn chứng minh.

Chữ đẹp nhỉ